Physics - Modern Physics Question with Solution | TestHub

According to photoelectric equation,

$hv-{\omega }_0=\frac{1}{2} m{v}_{max}^2$

$\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=\frac{1}{2} m{v}_{max}^2$

$hc\left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)=\frac{1}{2}m{v}_{max}^2$

$v_{max}=\sqrt{\frac{2hc}{m} \left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)}$

When wavelength is$\lambda$and velocity v, then

$v=\sqrt{\frac{2hc}{m} \left(\frac{{\lambda }_0-\lambda }{\lambda \cdot {\lambda }_0}\right)}$..... (i)

When wavelength is$\frac{3\lambda }{4}$and velocity is$v\prime$then

$v^{\prime }=\sqrt{\frac{2hc}{m}\left[\frac{{\lambda }_0-\left(\frac{3\lambda }{4}\right)}{\left(\frac{3\lambda }{4}\right)\times {\lambda }_0}\right]}$...... (ii

Divide equation (ii) by equation (i), we get

$\frac{v^{\prime }}{v}=\sqrt{\frac{{\lambda }_0-\frac{3\lambda }{4}}{\frac{3\lambda {\lambda }_0}{4}}\times \frac{\lambda {\lambda }_0}{{\lambda }_0-\lambda }}$

$v^{\prime }=v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{\left[{\lambda }_0-\left(\frac{3\lambda }{4}\right)\right]}{\lambda {\lambda }_0}}=v^{\prime }>v{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$